What is voltage drop and how is it calculated?

Voltage drop is the reduction in voltage between a source and its load caused by the supply wiring and connections. A power supply may read 24 V at its terminals while a distant device receives less. Excessive drop can affect operation even when the source itself is working correctly.

Separate voltage drop from power loss

Current flowing through cable resistance produces a voltage drop. For a simple resistive DC circuit, ΔV = I × R. The heat dissipated in the wiring is P_loss = I² × R = I × ΔV. Voltage is measured in volts; power loss is measured in watts.

The voltage–current distinction matters when load demand changes. With resistance held constant, doubling current doubles the voltage drop and quadruples the heating loss. The relationship between voltage and power explains why a cable can both reduce the voltage available to equipment and waste energy.

What makes the drop larger?

Longer conductors have more resistance, which is why long cable runs need careful assessment. Smaller cross-sections also increase resistance. A cable sizing decision must account for current and distance together.

Material and temperature change the result. Copper and aluminium conductors do not have the same resistance at the same cross-section. Copper resistance increases as temperature rises, so a cold-conductor estimate can understate the operating drop. Cable heating and voltage drop are therefore connected.

Contact resistance adds to cable resistance. Loose or damaged terminals and joints, or unsuitable lugs and crimps, may cause a concentrated drop. Installing a larger cable does not automatically repair a poor connection.

For DC, calculate the complete circuit loop

For a two-wire DC circuit with identical outgoing and return conductors, R_loop = 2 × ρ × L / S, giving ΔV = 2 × ρ × L × I / S. Here L is the one-way source-to-load distance, S is each conductor's cross-section, I is operating current and ρ is resistivity.

The examples use metres for L, mm² for S and Ω·mm²/m for ρ. An approximate copper value at 20 °C is 0.0175 Ω·mm²/m. Treat this as an educational estimate; use manufacturer resistance data and operating temperature for design. Convert AWG sizes to the appropriate cross-section before calculating.

If the length already includes both outgoing and return conductors, do not double it again. If their cross-sections differ, calculate each resistance separately and add them. Include connection resistance in the total loop resistance.

Worked example: a 24 V load 20 metres away

Assume a 24 V source, 5 A operating current, a 20 m one-way run and 1.5 mm² conductors. The current travels through 40 m of conductor in total. Ignoring contact resistance, R_loop = 2 × 0.0175 × 20 / 1.5 ≈ 0.467 Ω.

Voltage drop: 5 × 0.467 ≈ 2.33 V. Load voltage: 24 − 2.33 ≈ 21.67 V. Percentage drop: 100 × 2.33 / 24 ≈ 9.72%. Total resistive wiring loss is approximately 11.67 W. Whether the load operates correctly depends on its specified input voltage range.

Keeping the same current and distance, compare 1.5 mm², 2.5 mm² and 4 mm² conductors:

This comparison assumes that current remains 5 A; it does not approve a cable size. Current-carrying capacity, installation method, ambient temperature, protection and short-circuit withstand require separate checks. A constant-power electronic load may draw more current as voltage falls, so its operating point must be reassessed.

AC calculations need additional information

For sinusoidal steady-state operation with an inductive load, approximate voltage-magnitude drop is ΔV ≈ 2 × I × L × (r × cosφ + x × sinφ) for a single-phase two-conductor circuit. For a balanced three-phase circuit, the line-to-line drop is ΔV ≈ √3 × I × L × (r × cosφ + x × sinφ).

Here r and x are each conductor's resistance and reactance per unit length; L is the one-way length and units must match. Use line-to-line voltage as the three-phase percentage reference. Distorted, unbalanced and drive-output circuits need an appropriate calculation method.

Test the circuit while the load is operating

The basic comparison is source-terminal voltage versus load-terminal voltage under load. Use a suitable multimeter and voltage measurement method. On a DC circuit, assess the positive feed and return separately. Measuring across a section carrying normal operating current reveals that section's drop.

A correct no-load reading does not prove that the wiring is healthy. Little current means little resistive drop. Distinguish load-related low voltage from the source's own voltage fluctuations. An ammeter or suitable current clamp helps establish demand. Live testing requires appropriate equipment and competence; isolate the circuit before repairing connections.

Practical ways to reduce the drop

Increase conductor cross-section, shorten the route, use appropriate terminations and distribute high-current loads through suitable supply paths. A correctly designed local supply may help with distant control equipment. Avoid treating one percentage as a universal limit: applicable installation requirements and the equipment's minimum input voltage both matter.